Question

2sin3x
$> - sqrt{2}$
​

Answer 1

$sin3x>-frac{sqrt{2}}{2}$

Тогда:

$-frac{pi}{4}+pi k<3x<frac{5pi}{4}+pi k$, k=0,1,2...

Итого:

$-frac{pi}{12}+frac{pi k}{3}<x<frac{5pi}{12}+frac{pi k}{3}$